Question

A simple random sample of 60 items resulted in a sample mean of 63. The population standard deviation is 12.

a. Compute the 95% confidence interval for the population mean (to 1 decimal).

(  ,   )

b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).

(  ,   )

Answer 1

since population standard deviation is given so we use z-value to find confidence interval

(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)

95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)

(a) answer is (60,66)

95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)=63±1.96*12/sqrt(60)=63±3.04=

=(59.96,66.04)=(60,66) ( one decimal place approximation)

n=	60
sample mean=	63
s=	12.00
	z-value	margin of error	lower limit	upper limit
95% confidence interval	1.96	3.04	59.96	66.04

(b) answer is (60.85,65.15)

95% confidence interval for population mean=sample mean±z(0.05/2)*sd/sqrt(n)=63±1.96*12/sqrt(120)=63±2.15=

=(60.85,65.15) ( two decimal place approximation)

n=	120
sample mean=	63
s=	12.00
	z-value	margin of error	lower limit	upper limit
95% confidence interval	1.96	2.15	60.85	65.15